結果

問題 No.3346 Tree to DAG
コンテスト
ユーザー n_bitand_n_per_3
提出日時 2025-10-29 18:00:22
言語 C++23
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 591 ms / 2,000 ms
コード長 4,179 bytes
コンパイル時間 8,318 ms
コンパイル使用メモリ 455,968 KB
実行使用メモリ 12,052 KB
最終ジャッジ日時 2025-11-13 21:06:04
合計ジャッジ時間 10,603 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge4
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ファイルパターン 結果
sample AC * 2
other AC * 39
権限があれば一括ダウンロードができます

ソースコード

diff # 

#include <bits/stdc++.h>
#include <boost/multiprecision/cpp_int.hpp>
using namespace std;
using ll = long long;
const ll MOD = 998244353;
using boost::multiprecision::cpp_int;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int N;
    if (!(cin >> N)) return 0;
    vector<vector<int>> g(N+1);
    for (int i = 0; i < N-1; ++i) {
        int u,v; cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
    }

    // precompute pow2 (mod)
    int MAXP = N + 10;
    vector<ll> pow2(MAXP+1);
    pow2[0] = 1;
    for (int i = 1; i <= MAXP; ++i) pow2[i] = (pow2[i-1] * 2) % MOD;

    // Build parent and order (DFS iterative)
    vector<int> parent(N+1, -1);
    vector<int> order;
    order.reserve(N);
    stack<int> st;
    st.push(1);
    parent[1] = 0;
    while (!st.empty()) {
        int u = st.top(); st.pop();
        order.push_back(u);
        for (int v : g[u]) if (parent[v] == -1) {
            parent[v] = u;
            st.push(v);
        }
    }
    // down dp
    vector<int> down(N+1, 0);
    for (int i = (int)order.size()-1; i >= 0; --i) {
        int u = order[i];
        int best = 0;
        for (int v : g[u]) if (v != parent[u]) {
            best = max(best, 1 + down[v]);
        }
        down[u] = best;
    }

    // up dp
    vector<int> up(N+1, 0);
    for (int idx = 0; idx < (int)order.size(); ++idx) {
        int u = order[idx];
        int mx1 = -1, mx2 = -1;
        for (int v : g[u]) if (v != parent[u]) {
            int val = 1 + down[v];
            if (val > mx1) { mx2 = mx1; mx1 = val; }
            else if (val > mx2) { mx2 = val; }
        }
        for (int v : g[u]) if (v != parent[u]) {
            int use = mx1;
            int valv = 1 + down[v];
            if (valv == mx1) use = mx2;
            int cand = up[u];
            if (use > cand) cand = use;
            up[v] = (cand == -1 ? 0 : cand + 1);
        }
    }

    auto modnorm = [&](ll x)->ll{
        x %= MOD;
        if (x < 0) x += MOD;
        return x;
    };

    // We will keep the best value in two ways:
    // - best_mod: the value f mod MOD (to output)
    // - best_big: the exact value f as a big integer (for correct comparison)
    cpp_int best_big = 0;
    ll best_mod = 0;
    bool best_set = false;

    // For each node r, consider its components and take the top 3 distances x,y,z
    for (int r = 1; r <= N; ++r) {
        vector<int> comps;
        comps.reserve(g[r].size() + 1);
        comps.push_back(0);
        if (parent[r] != 0) comps.push_back(up[r]);
        for (int v : g[r]) if (v != parent[r]) {
            comps.push_back(1 + down[v]);
        }
        if ((int)comps.size() < 3) continue;
        int a=-1,b=-1,c=-1;
        for (int v : comps) {
            if (v > a) { c = b; b = a; a = v; }
            else if (v > b) { c = b; b = v; }
            else if (v > c) { c = v; }
        }
        int x = a, y = b, z = c;
        int s = x + y + z;

        // compute g = 2^{s+3} - (2^{x+2}+2^{y+2}+2^{z+2}) + 6  (exact integer)
        // then f = g * 2^{N-1-s}
        // We'll compute exact f as big integer for correct comparison, and also compute f mod MOD to output later.

        // compute g modulo MOD (for modular f)
        ll term1 = pow2[s+3];
        ll term2 = (pow2[x+2] + pow2[y+2]) % MOD;
        term2 += pow2[z+2]; term2 %= MOD;
        ll gmod = modnorm(term1 - term2 + 6);
        int exp = (N-1) - s;
        if (exp < 0) continue; // invalid, skip
        ll fmod = (pow2[exp] * gmod) % MOD;

        // compute exact f as big integer using cpp_int
        // g_big = (1 << (s+3)) - (1 << (x+2)) - (1 << (y+2)) - (1 << (z+2)) + 6
        cpp_int g_big = cpp_int(1);
        g_big <<= (s + 3);
        g_big -= (cpp_int(1) << (x + 2));
        g_big -= (cpp_int(1) << (y + 2));
        g_big -= (cpp_int(1) << (z + 2));
        g_big += 6;

        if (g_big < 0) continue; // safety, though shouldn't happen

        cpp_int f_big = g_big << exp;

        if (!best_set || f_big > best_big) {
            best_set = true;
            best_big = f_big;
            best_mod = fmod;
        }
    }

    cout << (best_mod % MOD) << "\n";
    return 0;
}
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